Sampling Practice Questions

Given that, $x_{1}(t)$ and $x_{2}(t)$ are two bandlimited signals having bandwidth $B=4 \pi \times 10^{3} \mathrm{rad} / \mathrm{sec}$ . and $y(t)=x_{2}(t) \cos \left(12 \pi \times 10^{3} t\right)+x_{1}(t) \cos \left(4 \pi \times 10^{3} t\right)$ So, \text{Nyquist Rate }=2\omega _{max}=2\[16 \pi × 10^3]$=32 \pi \times 10^3; rad/sec$

$\mathrm{NR}=2 \times f_{\mathrm{max}}=2 \times 15=30 \mathrm{kHz}$

Let us assume an arbitrary spectrum for x(t) as shown below: The spectrum of the sampled signal can be given as, For proper reconstruction of the signal,

The spectrum of the sampled signal can be given as shown below: So, three sinusoids will be there at the output of the LPF and the frequencies of those sinusoids are 2 kHz, 7 kHz and 11 kHz.

If x(t) is a message signal and y(t) is a sampled signal, then y(t) is related to x(t) as $y(t)=x(t) \sum_{n=-\infty }^{\infty } \delta\left(t-n T_{s}\right)$ $r(f)=f_{s} \sum_{n=--\infty }^{\infty } X\left(f-n f_{s}\right)$ Spectrum of X(f) and Y(f) are as shown Cut off frequency of LPF = 23 Hz Hence, frequency at the output is 13 Hz

After sampling x(t) by 15 Hz, we get $X_{s}(f)=15 \sum_{n=-\infty }^{\infty } X(f-15 n)$ Given, $\quad h(t)=\left[\frac{\sin (\pi t)}{\pi t}\right] \cos \left(40 \pi t-\frac{\pi}{2}\right)$ $H(f)=\frac{1}{2}\left[\text{rect} \frac{(f-20)}{2 \pi}+\text{rect} \frac{(f+20)}{2 \pi}\right] e^{-j \frac{\pi}{2}}$

$x(t)=\frac{\sin \left(\frac{\pi t}{2}\right)}{\left(\frac{\pi t}{2}\right)} * \sum_{n=-\infty }^{\infty } \delta[t-10 n]$

We know that

From equation (ii) $t \propto C$ and it is clear that if value of capacitor increases then the acquisition time increases. For capacitor the drop rate is given as $\frac{d v}{d t}$

From above relation it is clear that if capacitor value increases drop rate decreases.

Multiplication in time domain = convolution in frequency domain. $x_{1}(t) \cdot x_{2}(t) \leftrightarrow X_{1}(\omega) * X_{2}(\omega)$ So, highest frequency component contained by the convolved signal $z(t)=1500 \mathrm{Hz}$ $\therefore$ Nyquist rate $=2 \times 1500$ $=3000 \mathrm{Hz}=3 \mathrm{kHz}$

$x(t)=\cos (10 \pi t)+\cos (30 \pi t)$ Given sampling frequency, $f_{s}=20 \mathrm{Hz}$ $\omega_{s}=40 \pi \mathrm{rad} / \mathrm{sec}$ $Y(\omega)=\sum_{K=-\infty }^{\infty } X\left(\omega-K \omega_{s}\right)$ After sampling waveform will be Applying an ideal low-pass filter of cut-off frequency of $20 \mathrm{Hz}$ or $40 \pi \; \mathrm{ rad} / \mathrm{sec},$ we get the frequencies in reconstructed signal as $10 \pi \text{ and } 30 \pi \; \mathrm{rad} / \mathrm{sec}$ . or $5 \mathrm{Hz}$ and $15 \mathrm{Hz}$

So, at steady state y remains zero for all sampling frequencies $\omega_{S}$

$m(t) g(t) \leftrightarrow M(f) * G(f)$ After low pass filtering with $f_{c} = 1 kHz$ Output is zero.

Available frequency components are $n f_{s} \pm f_{m^{\prime}}$ So. $1 \mathrm{kHz}, 2.5 \mathrm{kHz}, 0.5 \mathrm{kHz} \ldots \ldots$ LPF has $f_{c}=0.8 \mathrm{k}$ only 0.5 k will appear at output.

From Fourier series expansion, $C_{n}=\frac{1}{T_{0}} \int_{-T_{0} / 6}^{T_{0} / 6} A \cdot e^{-j \pi \omega_{0} t} d t=\frac{A}{\pi n} \sin \left(\frac{n \pi}{3}\right)$ $\therefore$ From $C_{n}$ its clear that $1,2,4,5,7 \ldots$ harmonics are present. Freq of p(t) corr. to $1,2,4,5,7\ldots$ are $10^{3} , 2 \times 10^{3}, 4 \times 10^{3} \ldots$ frequency components $0.7 \mathrm{k}$ and $0.4 \mathrm{k}$ $(t)^{(t)} h a s^{\prime}(t) \operatorname{gives}(1 \pm 0.7) k,(2 \pm 0.7) k$ $(4 \pm 0.7) k(1 \pm 0.4) k,(2 \pm 0.4) k \ldots$ Frequency present in range of $2.5 \mathrm{k}$ to $3.5 \mathrm{k}$ are $2.7, 3.3$